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http://numericalmethods.eng.usf.edu 1 Newton’s Divided Difference Polynomial Method of Interpolation Mechanical Engineering Majors Authors: Autar Kaw, Jai Paul http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates
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Newton’s Divided Difference Method of Interpolation http://numericalmethods.eng.usf.edu http://numericalmethods.eng.usf.edu
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3 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1 ), …… (x n,y n ), find the value of ‘y’ at a value of ‘x’ that is not given.
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http://numericalmethods.eng.usf.edu4 Interpolants Polynomials are the most common choice of interpolants because they are easy to: Evaluate Differentiate, and Integrate.
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http://numericalmethods.eng.usf.edu5 Newton’s Divided Difference Method Linear interpolation: Given pass a linear interpolant through the data where
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http://numericalmethods.eng.usf.edu6 Example A trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for linear interpolation. Temperature ( o F) Thermal Expansion Coefficient (in/in/ o F) 806.47 × 10 −6 06.00 × 10 −6 − 60 5.58 × 10 −6 − 160 4.72 × 10 −6 − 260 3.58 × 10 −6 − 340 2.45 × 10 −6
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http://numericalmethods.eng.usf.edu7 Linear Interpolation
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http://numericalmethods.eng.usf.edu8 Linear Interpolation (contd)
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http://numericalmethods.eng.usf.edu9 Quadratic Interpolation
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http://numericalmethods.eng.usf.edu10 Example A trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for quadratic interpolation. Temperature ( o F) Thermal Expansion Coefficient (in/in/ o F) 806.47 × 10 −6 06.00 × 10 −6 − 60 5.58 × 10 −6 − 160 4.72 × 10 −6 − 260 3.58 × 10 −6 − 340 2.45 × 10 −6
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http://numericalmethods.eng.usf.edu11 Quadratic Interpolation (contd)
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http://numericalmethods.eng.usf.edu12 Quadratic Interpolation (contd)
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http://numericalmethods.eng.usf.edu13 Quadratic Interpolation (contd)
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http://numericalmethods.eng.usf.edu14 General Form where Rewriting
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http://numericalmethods.eng.usf.edu15 General Form
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http://numericalmethods.eng.usf.edu16 General form
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http://numericalmethods.eng.usf.edu17 Example A trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the direct method for cubic interpolation. Temperature ( o F) Thermal Expansion Coefficient (in/in/ o F) 806.47 × 10 −6 06.00 × 10 −6 − 60 5.58 × 10 −6 − 160 4.72 × 10 −6 − 260 3.58 × 10 −6 − 340 2.45 × 10 −6
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http://numericalmethods.eng.usf.edu18 Example
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http://numericalmethods.eng.usf.edu19 Example
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http://numericalmethods.eng.usf.edu20 Example
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http://numericalmethods.eng.usf.edu21 Comparison Table
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http://numericalmethods.eng.usf.edu22 Reduction in Diameter The actual reduction in diameter is given by where T r = room temperature (°F) T f = temperature of cooling medium (°F) Since T r = 80 °F and T r = −108 °F, Find out the percentage difference in the reduction in the diameter by the above integral formula and the result using the thermal expansion coefficient from the cubic interpolation.
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http://numericalmethods.eng.usf.edu23 Reduction in Diameter We know from interpolation that Therefore,
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http://numericalmethods.eng.usf.edu24 Reduction in diameter Using the average value for the coefficient of thermal expansion from cubic interpolation The percentage difference would be
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Additional Resources For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://numericalmethods.eng.usf.edu/topics/newton_div ided_difference_method.html
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THE END http://numericalmethods.eng.usf.edu
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